[Big-Daddy]

Estimate the mole fraction x(Ar) of argon in Rayleigh's atmospheric nitrogen, by assuming that argon and nitrogen were the only constituents. The volume of the flask in which the atmospheric nitrogen held is 1.940394518 * 10^-3 m³. The pressure of the flask is atmospheric pressure. When the flask is filled with pure nitrogen and has no argon, it contains 2.2990 g of nitrogen; when it is filled with atmospheric nitrogen (containing traces of argon), it contains 2.3102 g of gas.

Comments:
I don't see why the information about the pure nitrogen is required? Can't we just say that, for the atmospheric nitrogen container,

n(Ar) * M_r(Ar) + n(N₂) * M_r(N₂) = m(total)

Then from the ideal gas equation

( p(Ar) * V * M_r(Ar) + p(N₂) * V * M_r(Ar) ) / (R * T) = m(total)

And then using the replacements p(Ar) = p(total) * x(Ar) and p(N₂) = p(total) * (1 - x(Ar)), we get an equation we can rearrange for x(Ar). This gives the same answer as the book's solution (0.01142 i.e. roughly 1.14%). So why does it bother providing all this information about pure chemical nitrogen, when, given the temperature, total pressure and volume, we can calculate the mole fraction and partial pressure of each component with 2 unknowns (2 unknown mole fractions and partial pressures) without any of this extra info?